$\overline{AC} = 3$ $\overline{BC} = {?}$ $A$ $C$ $B$ $3$ $?$ $ \sin( \angle BAC ) = \frac{3\sqrt{10} }{10}, \cos( \angle BAC ) = \frac{ \sqrt{10}}{10}, \tan( \angle BAC ) = 3$
Solution: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{\overline{BC}}{3} $ $ \overline{BC}=3 \cdot \tan( \angle BAC ) = 3 \cdot 3 = 9$